Talk:Infinite product

This article is probably very non-ideal, but when I needed this material a while ago it was hard to find, and so I figured wikipedia would be a good place to hold it. Please feel free to change it around and make it more encyclopedia-like, even more so than usual. :)

We probably need a little caveat about negative numbers: the log formula won't work without problems in that case. Maybe use a different log which is defined on the negative reals? --AxelBoldt
 * A product of negative coefficients does not converge; it can at most flip. --Yecril (talk) 18:00, 20 December 2008 (UTC)

Inaccurate Statement
"Therefore, the logarithm log an will be defined for all but a finite number of n" neither makes sense nor true as I understand it. One can easily imagine a sequence which converges to 1 that is not finite. one simple example is a(n)= 1+(1/n) where n is an integer. This of course is not limited for a being real, for example think of any oscillating function which dampens to 1. this might have been a simple typo the author made, but i don't understand his/her intention, so i am timid to edit this —Preceding unsigned comment added by 129.49.61.161 (talk) 17:06, 17 August 2010 (UTC)


 * Maybe it is a left-over from careless editing. Some sources allow a finite number of 0 factors and still call the product convergent if the product of the non-zero factors is non-zero.  Probably that is the meaning. I'll try to fix it. McKay (talk) 06:09, 31 March 2011 (UTC)

Why is $$\prod_{k=1}^{\infty} a_k = \lim_{n\rightarrow\infty} \prod_{k=1}^{n} a_k = 0$$ an example of divergence?
The article currently indicates that if the infinite product is zero then that is an example of divergence. Why? I would think that that is an example of convergence, despite that a corresponding sum $$\sum_{k=1}^{\infty} \log(a_k)$$ is not finite. — Q uantling (talk &#124; contribs) 18:35, 30 March 2011 (UTC)


 * It is just a convention. The reason for the convention is that allowing convergence to 0 would admit too many uninteresting series (for example, any sequence {an} with | an| &lt; 1/2), and some desirable properties would disappear. McKay (talk) 06:15, 31 March 2011 (UTC)
 * Perhaps it's worth nothing that following that convention, we also neglect many "interesting" examples. Say $$0<a_1<1$$. Can $$a_k$$ converge to 1, and still result in the product converging to 0? If not, then what is the highest number $$a_k$$ can approach? Forgive me if these are well known and simple problems to solve, it's new territory for me.
 * Why should it be deemed that non-trivially converging-to-0 infinite products are necessarily uninteresting? 2601:647:C900:B6C0:99ED:F4A6:FA90:CDFA (talk) 05:52, 16 December 2022 (UTC)

How does this fit in?
If F(x) can be represented as a product:


 * $$F(x)=\Pi_{n=1}^\infty f_n(x)$$

then define G(x)=1-F(x) and using the series expansion for the natural logarithm: $$\ln(1-G(x))= -\sum_{n=1}^\infty G(x)^n/n $$


 * $$\ln(F(x))=-\sum_{n=1}^\infty G(x)^n/n = \sum_{n=1}^\infty \ln(f_n(x))$$

so that $$f_n(x)=e^{-G(x)^n/n}$$ and


 * $$F(x)=\Pi_{n=1}^\infty\, e^{-\frac{1}{n}(1-F(x))^n}$$

I have checked this on Mathematica for a number of functions, and Mathematica agrees, but I have no idea what the conditions are for this formula to work. Can anyone find a reference for this and put a section into the article? Its a very useful and general expression, and much simpler than the more general Weierstrass development. PAR (talk) 17:45, 13 June 2011 (UTC)


 * Ok, probably after almost 6 years this comment might have been forgotten, but let me explain why I don't think is a good idea to include such observation in the article. What you say is a trivial statement: if $$c \in [-1,1]$$ is a fixed number, then


 * $$\Pi_{n=1}^\infty\, e^{-\frac{c^n}{n}} = \exp(-\sum_{n=1}^\infty c^n/n) = \exp(\log(1-c)) = 1-c. $$


 * Since you are expanding $$\log(1-F(x))$$, you are assuming that $$F(x) \in [-1,1]$$ for every $$x$$, and then you are just applying the above formula point-wise. Note that this is quite different from Weierstrass' formula (if you wish, the problem with your expression is that you write $$F(x)$$ in terms of $$F(x)$$, so that is not very useful, just like is not very useful to say that -for example- $$F(x) = [2 F(x)] * \left[\frac{F(x)}{2}\right]$$). Lucha (talk) 16:10, 28 February 2017 (UTC)